Calculate Impedance Magnitude & Phase of VR

[Zell2]

Homework Statement

There is a circuit with an alternating voltage source and a capacitor (capacitance C) an inductor (inductance L) and a resistor (resistance R). The resistor has a potential VR marked across it.

Calculate the magnitude and phase of the voltage VR.

Homework Equations

Impedance of resistor=R, impedance of capacitor=-j/wC, impedance of inductor=jwL, impendances in series add linearly.

The Attempt at a Solution

My first though was just to treat it as a potential divider and its easy to find the magnitude. But I'm getting at bit confused about finding the phase of VR: can you just take the phase from the potential divider equation, so pahse relative to input voltage=-arctan((wL-1/wC)/R)? Possibly it's me trying to make things harder than they should be but it's from a past exam paper and was worth 12 marks so doesn't seem to involve much work?
Thanks

 

[BobG]

Homework Statement

There is a circuit with an alternating voltage source and a capacitor (capacitance C) an inductor (inductance L) and a resistor (resistance R). The resistor has a potential VR marked across it.

Calculate the magnitude and phase of the voltage VR.

Homework Equations

Impedance of resistor=R, impedance of capacitor=-j/wC, impedance of inductor=jwL, impendances in series add linearly.

The Attempt at a Solution

My first though was just to treat it as a potential divider and its easy to find the magnitude. But I'm getting at bit confused about finding the phase of VR: can you just take the phase from the potential divider equation, so pahse relative to input voltage=-arctan((wL-1/wC)/R)? Possibly it's me trying to make things harder than they should be but it's from a past exam paper and was worth 12 marks so doesn't seem to involve much work?
Thanks

Yes, except I don't know why you have negative arctan. Add the impedance of the inductor to the impedance of the capacitor (which is negative). The result gives you a net inductance or net capacitance.

 

[Zell2]

The negative sign came from:
VR=RV/(R+jwL-j/(wC))
so I thought:
phase(VR)=-phase(R+jwL-j/(wC))
Is this right?
Thanks

 

[berkeman]

The negative sign came from:
VR=RV/(R+jwL-j/(wC))
so I thought:
phase(VR)=-phase(R+jwL-j/(wC))
Is this right?
Thanks

If you were using complex exponential form for the complex numbers, then yes, $$\frac{1}{e^x} = e^{-x}$$

But with the complex number in algebraic format like that, I don't think you can make the same assumption. It may be true, but I'd work it out first to be sure.

And your problem statement wasn't very clear. You have a series C-L-R circuit with the components in that order, and the phase you want is the phase shift from the driving source (which drives the top of the C with respect to the bottom of the R)? You need to be careful with the definition of "phase" in this problem. There is no phase shift across the resistor, obviously, so the "phase" that is being asked for has to be with respect to the driving source. And in that case, the order of the components makes a difference.

 

[Zell2]

You have a series C-L-R circuit with the components in that order,

Yes that's the circuit.

and the phase you want is the phase shift from the driving source (which drives the top of the C with respect to the bottom of the R)?

I need the phase of the voltage across the resistance, (no direction specified), with respect to the voltage source. Sorry about the lack of clarity in the first post.

I'm a bit confused about the direction. The voltage across the resistor in the direction opposing the source voltage will the in phase with the current, and Vcomponents=(R+jwL-j/wc)I. but do I need the voltage in this direction or the opposite one?

Thanks

 

[berkeman]

I would interpret the question as asking for the phase of the voltage waveform at the top of the resistor as compared to the zero phase of the driving source.